2Classical gases

II Statistical Physics 2.5 Interacting gases
So far, we have been talking about ideal gases. What happens if there are
interactions?
For a real gas, if they are sufficiently dilute, i.e.
N/V
is small, then we expect
the interactions to be negligible. This suggests that we can try to capture the
effects of interactions perturbatively in
N
V
. We can write the ideal gas law as
p
kT
=
N
V
.
We can think of this as a first term in an expansion, and add higher order terms
p
kT
=
N
V
+ B
2
(T )
N
2
V
2
+ B
3
(T )
N
3
V
3
+ ··· .
Note that the coefficients depend on
T
only, as they should be intensive quantities.
This is called the Virial expansion, and the coefficients
B
k
(
T
) are the Virial
coefficients. Our goal is to figure out what these B
k
(T ) are.
We suppose the interaction is given by a potential energy
U
(
r
) between two
neutral atoms (assuming monoatomic) at separation r.
Example. In genuine atoms, for large r (relative to atomic size), we have
U(r)
1
r
6
.
This comes from dipole-dipole interactions. Heuristically, we can understand
this power of 6 as follows while the expectation values of electric dipole of an
atom vanishes, there exists non-trivial probability that the dipole
p
1
is non-zero.
This gives an electric field of
E
p
1
r
3
.
This induces a dipole p
2
in atom 2. So we have
p
2
E
p
1
r
3
.
So the resulting potential energy is
U p
2
E
p
2
1
r
6
.
This is called the van der Waals interaction. Note that the negative sign means
this is an attractive force.
For small
r
, the electron orbitals of the atoms start to overlap, and then we get
repulsion due to the Pauli principle. All together, we obtain the Lennard-Jones
potential given by
U(r) = U
0
r
0
r
12
r
0
r
6
.
r
U(r)
To make life easy, we will actually use a “hard core repulsion” potential instead,
given by
U(r) =
(
r < r
0
U
0
r
0
r
6
r > r
0
r
U(r)
r
0
For a general U, we can write the Hamiltonian of the gas is
H =
N
X
i=1
p
2
i
2m
+
X
i>j
U(r
ij
),
with
r
ij
= |r
i
r
j
|.
is the separation between particle i and particle j.
Because of this interaction, it is no longer the case that the partition function
for
N
particles is the
N
th power of the partition function for one particle. We
have
Z(N, V, T ) =
1
N
1
(2π~)
3n
Z
N
Y
i=1
d
3
p
i
d
3
r
i
e
βH
=
1
N!
1
(2π~)
3n
Z
Y
i
d
3
p
i
e
βp
2
i
/2m
!
Z
Y
i
d
3
r
i
e
β
P
j<k
U(r
jk
)
!
=
1
N!λ
3N
Z
Y
i
d
3
r
i
e
β
P
j<k
U(r
jk
)
,
where again
λ =
r
2π~
2
mkT
.
Since we want to get an expansion, we might want to expand this in terms of
the potential, but that is not helpful, because the potential is infinite for small
r
.
Instead it is useful to consider the Mayer f-function:
f(r) = e
βU(r)
1.
This function has the property that
f
(
r
) =
1 for
r < r
0
(in the case of the
hardcore repulsion), and
f
(
r
)
0 as
r
. So this is a nicer function as it
only varies within this finite range.
We further simplify notation by defining
f
ij
= f(r
ij
).
Then we have
Z(N, V, T ) =
1
N!λ
3N
Z
Y
i
d
3
r
i
Y
j<k
(1 + f
jk
)
=
1
N!λ
3N
Z
Y
i
d
3
r
i
1 +
X
j<k
f
jk
+
X
j<k
X
`<m
f
jk
f
`m
+ ···
.
The first term is just
Z
Y
i
d
3
r
i
= V
N
,
and this gives the ideal gas term. Now each of the second terms is the same, e.g.
for j = 1, k = 2, this is
Z
Y
i
d
3
r
i
f
12
= V
N2
Z
d
3
r
1
d
3
r
2
f
12
= V
N1
I,
where we set
I =
Z
d
3
r f (r).
Since
f
(
r
)
0 as
r
, we might as well integrate over all space. Summing
over all terms, and approximating
N
(
N
1)
/
2
N
2
/
2, we find that the first
two terms of the partition function are
Z(N, V, T ) =
V
N
N!λ
3N
1 +
N
2
2V
I + ···
.
Up to first order, we can write this as
Z(N, V, T ) =
V
N
N!λ
3N
1 +
N
2V
I + ···
N
= Z
ideal
1 +
N
2V
I + ···
N
.
This pulling out of the
N
to the exponent might seem a bit arbitrary, but writing
it this way, it makes it much clearer that
S
,
F
etc would be extensive quantities.
For example, we can write down the free energy as
F = kT log Z = F
ideal
NkT log
1 +
N
2V
I + ···
.
Without actually computing
I
, we can expect that it grows as
I r
3
0
. Since we
are expanding in terms of NI/V , we need
N
V
1
r
3
0
.
So we know the expansion is valid if the density of the gas is much less than the
density of an atom. In real life, to determine the density of an atom, we need to
find a system where the atoms are closely packed. Thus, it suffices to measure
the density of the substance in liquid or solid form.
Assuming that the gas is indeed not dense, we can further use the approxi-
mation
log(1 + x) x.
So we have
p =
F
V
T
=
NkT
V
1
N
2V
I + ···
.
So we have
pV
NkT
= 1
N
2V
I + ··· .
So we can now read off what the second Virial coefficient is:
B
2
=
1
2
I.
We can consider what we get with different potentials. If we have a completely
repulsive potential, then
U
(
r
)
>
0 everywhere. So
f <
0, and thus
B
2
(
t
)
>
0. In
other words, having a repulsive interaction tends to increase the pressure, which
makes sense. On the other hand, if we have an attractive potential, then the
pressure decreases.
If we have a hardcore repulsion, then we have
I =
Z
r=r
0
r=0
d
3
r (1) +
Z
r=r
0
d
3
r
e
βU
0
(r
0
/r)
6
1
.
The second term is slightly tricky, so we’ll restrict to the case of high temperature,
hence small β. We can write
e
βU
0
(r
0
/r)
6
1 + βU
0
r
0
r
6
+ ···
Then we get
I
4
3
πr
3
0
+
4πU
0
kT
Z
r
0
dr
r
6
0
r
6
=
4πr
3
0
3
U
0
kT
1
.
For large
T
, this is negative, and so the repulsive interaction dominates. We can
plug this into our equation of state, and find
pV
NkT
1
N
V
a
kT
b
,
where
a =
2πr
3
0
3
U
0
, b =
2πr
3
0
3
.
We can invert this to get
kT =
V
N
p +
N
2
V
2
a
1 +
N
V
b
1
.
Taking the Taylor expansion of the inverse and truncating higher order terms,
we obtain
kT =
p +
N
2
V
2
a
V
N
b
.
This is the van der Waals equation of state, which is valid at low density
(Nr
3
0
/V 1) and high temperature (βU
0
1).
We can also write this as
p =
NkT
V bN
a
N
2
V
2
.
In this form, we see that
a
gives a reduced pressure due to long distance attractive
force, while we can view the
b
contribution as saying that the atoms take up
space, and hence reduces the volume.
By why exactly the factor of
bN
? Imagine two atoms living right next to
each other.
r
0
The value
r
0
was chosen to be the distance between the two cores, so the volume
of each atom is
4π
3
r
0
2
3
,
which is not
b
. We might think the right way to do this problem is to look at
how much volume the atom excludes, i.e. the dashed volume. This is
Ω =
4
3
πr
3
0
= 2b.
This is again not right. So we probably want to think more carefully.
Suppose we have
N
particles. When we put in the first atom, the amount
of space available to put it is
V
. When we now try to put in the second, the
available space is
V
(assuming the two atoms are not too close together so
that they don’t exclude the same volume). Similarly, when we put in the third,
the available space is V 2Ω.
Diving by
N
! for indistinguishability, we find that the total phase space
volume available for placing our particles is
1
N!
V (V Ω)(V 2Ω) ···(V (N 1)Ω)
1
N!
V
N
1
N
2
2
V
+ ···
1
N!
V
N
2
N
,
which explains why the reduced volume for each particle is /2 instead of Ω.
Now suppose we want to find higher-order corrections to the partition function
in the Virial expansion. The obvious thing might be to take them to be the
contributions by
PP
f
jk
f
`m
. However, this is not quite right. If we try to keep
track of how large the terms are carefully, we will get a different answer.
The way to do it properly would be via the cluster expansion, which involves
using nice diagrams to figure out the right terms to include, similar to how
Feynman diagrams are used to do perturbation theory in quantum field theory.
Unfortunately, we do not have the time to go into the details.